I've read through the standard explanation of the electric field due to a spherical shell with uniform charge density. This explanation argues that because a Gaussian surface inside the shell encloses zero charge, by gauss' law, the electric field must be zero here. However, shouldn't the statement actually be that the flux is zero here? For example, a Gaussian sphere enclosing no charge above an infinite plane has zero net flux through it, and so it satisfies Gausss' theorem, but inside the sphere, the electric field is still the standard field due to an infinite sheet, no? So doesn't the argument re the spherical shell only rule out that the flux thru a gaussian surface inside the shell will be zero, and not that the field there is zero?

  • $\begingroup$ Isn't there another component to the argument that you've left out? $\endgroup$ – Alfred Centauri Sep 18 '19 at 23:13
  • $\begingroup$ ...namely spherical symmetry of system? $\endgroup$ – Qmechanic Sep 18 '19 at 23:52

You are right, Gauss's law in its integral form generally just relates the charge enclosed by the Gaussian surface to the flux of the electric field through our Gaussian surface

$$\oint\mathbf E\cdot\text d\mathbf a=\frac{Q_\text{encl}}{\epsilon_0}$$

i.e., if there is no enclosed charge, then we know there is no flux through the surface.

However, as pointed out in the comments, if we have chosen the Gaussian surface carefully such that symmetry allows us to argue that if there is a field, if must point perpendicularly to the surface and must have a constant magnitude on this surface, then it must be the case that $$\oint\mathbf E\cdot\text d\mathbf a=\oint E\ \text d a=E\oint\ \text d a=\frac{Q_\text{encl}}{\epsilon_0}$$

And so now we can make the argument that since $\oint\ \text d a\neq0$, if $Q_\text{encl}=0$ it must be that $E=0$.

  • $\begingroup$ I see the symmetry when the Gaussian surface/sphere is centered at the origin at the center of the spherical shell, so I see why E must equal zero at the center. However, I'm having a hard time seeing the symmetry when this is not the case, and the gaussian surface is centered elsewhere within the shell. $\endgroup$ – Marcel Mazur Sep 19 '19 at 2:15
  • $\begingroup$ Wait I think I understand now. The lack of symmetry in other cases does not matter. We can always center a gaussian sphere at the origin, and argue by symmetry that E must be the same everywhere on the surface. Then we can use your argument to show that E is zero. And we can do this for any radius inside the shell, so we can show that at any point on any concentric Gaussian surface in the shell, E must be zero. Is this right/what you are saying? $\endgroup$ – Marcel Mazur Sep 19 '19 at 2:40
  • $\begingroup$ @MarcelMazur Yes that is exactly right. What you have proposed is sufficient. $\endgroup$ – BioPhysicist Sep 19 '19 at 4:10

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